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Apr 28, 2017 at 10:03 history edited Qfwfq CC BY-SA 3.0
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Apr 28, 2017 at 9:55 history edited Francesco Polizzi CC BY-SA 3.0
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Apr 28, 2017 at 9:50 comment added Francesco Polizzi Whitney stratification still exist in the complex setting. See my edit to the answer.
Apr 28, 2017 at 9:49 history edited Francesco Polizzi CC BY-SA 3.0
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Apr 28, 2017 at 9:34 comment added user237522 Please, what happens if we replace $\mathbb{R}$ by $\mathbb{C}$?
Apr 27, 2017 at 18:32 comment added Francesco Polizzi @Libli: you are right. This does not define in general a Whitney stratification, but some refinement will, see J. L. Verdier, Stratifications de Whitney et theoreme de Bertini-Sard, Invent. Math. 36 (1976), 295-312. Thank you for the remark.
Apr 27, 2017 at 18:16 comment added Libli Note however that the wikipedia reference that you provide contains a big mistake. If $X$ is a singular variety, the collection $Sing(X), Sing(Sing(X)),...$ is not a Whitney stratification in general.
Apr 27, 2017 at 17:14 vote accept Arturo
Apr 27, 2017 at 15:14 history edited Francesco Polizzi CC BY-SA 3.0
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Apr 27, 2017 at 14:26 history answered Francesco Polizzi CC BY-SA 3.0